Physics, asked by Vishwakarm7724, 1 year ago

Potential energy of a particle at position x is given by u=x^2-5x

Answers

Answered by lidaralbany
8

Answer:

Potential is minimum at x=2.5

Minimum Potential = -6.25

Explanation:

Given that,

u = x^{2}-5x....(I)

On differentiate of u

\dfrac{du}{dx} = 2x-5

Now,

\dfrac{du}{dx} = 0

Then, 2x-5 = 0

2x = 5

x = \dfrac{5}{2}

x = 2.5

Now, double derivative is

\dfrac{d^{2}u}{dx^{2}} = 2

At x = 2.5, double derivative is positive.

So, put the value of x in equation (I)

u = (2.5)^{2}-5\times2.5

u = -6.25

Hence, potential is minimum at x=2.5 Minimum Potential = -6.25

Answered by mindfulmaisel
3

"x=2.5

As we know that a particle is said to be in equilibrium if net force acting on the particle is zero .

we can also go through this formula

{ F }_{ x }\quad =\quad \frac { du }{ dx }

For Force=0 we have to make \frac { dU }{ dX } \quad =\quad 0

now put the value in first derivative i.e 2X-5=0 .

x=2.5 ans"

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